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Q&A Why does `let map f = id >=> switch f` work in F#?

Asked How to implement map using the fish (>=>, Kleisli composition) operator in F#? a couple of hours ago, and r~~'s answer blew my mind: let map f = id >=> switch f It is perfect ...

1 answer · posted 9mo ago by toraritte‭ · edited 8mo ago by Alexei‭

Question f# functional-programming monads monadic-composition

#4: Post edited by

Alexei‭ · 2024-04-10T15:22:44Z (8 months ago)
added relevant tag

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functional-programming monads monadic-composition

f# functional-programming monads monadic-composition

#3: Post edited by

toraritte‭ · 2024-04-03T18:02:31Z (9 months ago)

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Asked [How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?](https://software.codidact.com/posts/291227) a couple of hours ago, and [r~~'s answer blew my mind](https://software.codidact.com/posts/291227/291228#answer-291228):
```
let map f = id >=> switch f
```
It is perfect in its simplicity, but when I look at the type signatures, it is not supposed to work. I've been at it for almost an hour now, so I'm probably missing something fundamental how F# evaluates expressions...
For the record, these are the implementations of `bind`,`switch`, and `>=>`:
```
let bind
( f : 'a -> Result<'b,'c>)
(result : Result<'a,'c>)
=
match result with
| Ok o -> f o
| Error e -> Error e
let switch
(f : 'a -> 'b)
(x : 'a )
=
f x |> Ok
let (>=>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
=
f >> (bind g)
```
My next attempt was using an alternative implementation for `>=>`:
```
let (>=>>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
x
=
match (f x) with
| Ok o -> g o
| Error e -> Error e
```
Here's the test invocation on `dotnet fsi`:
```
(id >=>> (switch ((+) 2) : int -> Result<int,string>))
((Ok 27) : Result<int,string>)
//=> Ok 29
```
I'm already stuck at why `>=>>` does not blow up on `id` as its first argument?
This is how I would think evaluation goes, but apparently this is not it:
```
id >=>> switch ((+) 2)
|
V
(>=>>) id (switch ((+) 2))
|
V
match (id x) with
| Ok o -> (switch ((+) 2)) o
| Error e -> Error e
```
---
Note to future self:...
```
(>=>>) (>=>>)
(f : 'a -> Result<'b,'error>) id
(g : 'b -> Result<'c,'error>) (Ok << ((+) 2) : int -> Result<int,string>))
x ((Ok 27) : Result<int,string>)
~~```~~

Asked [How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?](https://software.codidact.com/posts/291227) a couple of hours ago, and [r~~'s answer blew my mind](https://software.codidact.com/posts/291227/291228#answer-291228):
```
let map f = id >=> switch f
```
It is perfect in its simplicity, but when I look at the type signatures, it is not supposed to work. I've been at it for almost an hour now, so I'm probably missing something fundamental how F# evaluates expressions...
For the record, these are the implementations of `bind`,`switch`, and `>=>`:
```
let bind
( f : 'a -> Result<'b,'c>)
(result : Result<'a,'c>)
=
match result with
| Ok o -> f o
| Error e -> Error e
let switch
(f : 'a -> 'b)
(x : 'a )
=
f x |> Ok
let (>=>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
=
f >> (bind g)
```
My next attempt was using an alternative implementation for `>=>`:
```
let (>=>>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
x
=
match (f x) with
| Ok o -> g o
| Error e -> Error e
```
Here's the test invocation on `dotnet fsi`:
```
(id >=>> (switch ((+) 2) : int -> Result<int,string>))
((Ok 27) : Result<int,string>)
//=> Ok 29
```
I'm already stuck at why `>=>>` does not blow up on `id` as its first argument?
This is how I would think evaluation goes, but apparently this is not it:
```
id >=>> switch ((+) 2)
|
V
(>=>>) id (switch ((+) 2))
|
V
match (id x) with
| Ok o -> (switch ((+) 2)) o
| Error e -> Error e
```
---
Note to future self:...
```
(>=>>) (>=>>)
(f : 'a -> Result<'b,'error>) id
(g : 'b -> Result<'c,'error>) (Ok << ((+) 2) : int -> Result<int,string>))
x ((Ok 27) : Result<int,string>)
```
(... and make sure to convert a point-free function if it does not make sense at first.)

#2: Post edited by

toraritte‭ · 2024-04-03T10:10:43Z (9 months ago)

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Asked [How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?](https://software.codidact.com/posts/291227) a couple of hours ago, and [r~~'s answer blew my mind](https://software.codidact.com/posts/291227/291228#answer-291228):
```
let map f = id >=> switch f
```
It is perfect in its simplicity, but when I look at the type signatures, it is not supposed to work. I've been at it for almost an hour now, so I'm probably missing something fundamental how F# evaluates expressions...
For the record, these are the implementations of `bind`,`switch`, and `>=>`:
```
let bind
( f : 'a -> Result<'b,'c>)
(result : Result<'a,'c>)
=
match result with
| Ok o -> f o
| Error e -> Error e
let switch
(f : 'a -> 'b)
(x : 'a )
=
f x |> Ok
let (>=>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
=
f >> (bind g)
```
My next attempt was using an alternative implementation for `>=>`:
```
let (>=>>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
x
=
match (f x) with
| Ok o -> g o
| Error e -> Error e
```
Here's the test invocation on `dotnet fsi`:
```
(id >=>> (switch ((+) 2) : int -> Result<int,string>))
((Ok 27) : Result<int,string>)
//=> Ok 29
```
I'm already stuck at why `>=>>` does not blow up on `id` as its first argument?
This is how I would think evaluation goes, but apparently this is not it:
```
id >=>> switch ((+) 2)
|
V
(>=>>) id (switch ((+) 2))
|
V
match (id x) with
| Ok o -> (switch ((+) 2)) o
| Error e -> Error e
```

Asked [How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?](https://software.codidact.com/posts/291227) a couple of hours ago, and [r~~'s answer blew my mind](https://software.codidact.com/posts/291227/291228#answer-291228):
```
let map f = id >=> switch f
```
It is perfect in its simplicity, but when I look at the type signatures, it is not supposed to work. I've been at it for almost an hour now, so I'm probably missing something fundamental how F# evaluates expressions...
For the record, these are the implementations of `bind`,`switch`, and `>=>`:
```
let bind
( f : 'a -> Result<'b,'c>)
(result : Result<'a,'c>)
=
match result with
| Ok o -> f o
| Error e -> Error e
let switch
(f : 'a -> 'b)
(x : 'a )
=
f x |> Ok
let (>=>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
=
f >> (bind g)
```
My next attempt was using an alternative implementation for `>=>`:
```
let (>=>>)
(f : 'a -> Result<'b,'error>)
(g : 'b -> Result<'c,'error>)
x
=
match (f x) with
| Ok o -> g o
| Error e -> Error e
```
Here's the test invocation on `dotnet fsi`:
```
(id >=>> (switch ((+) 2) : int -> Result<int,string>))
((Ok 27) : Result<int,string>)
//=> Ok 29
```
I'm already stuck at why `>=>>` does not blow up on `id` as its first argument?
This is how I would think evaluation goes, but apparently this is not it:
```
id >=>> switch ((+) 2)
|
V
(>=>>) id (switch ((+) 2))
|
V
match (id x) with
| Ok o -> (switch ((+) 2)) o
| Error e -> Error e
```
---
Note to future self:...
```
(>=>>) (>=>>)
(f : 'a -> Result<'b,'error>) id
(g : 'b -> Result<'c,'error>) (Ok << ((+) 2) : int -> Result<int,string>))
x ((Ok 27) : Result<int,string>)
```

#1: Initial revision by

toraritte‭ · 2024-04-02T22:08:47Z (9 months ago)

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Why does `let map f = id >=> switch f` work in F#?

Asked [How to implement `map` using the fish (>=>, Kleisli composition) operator in F#?](https://software.codidact.com/posts/291227) a couple of hours ago, and [r~~'s answer blew my mind](https://software.codidact.com/posts/291227/291228#answer-291228):

```
let map f = id >=> switch f
```

It is perfect in its simplicity, but when I look at the type signatures, it is not supposed to work. I've been at it for almost an hour now, so I'm probably missing something fundamental how F# evaluates expressions...

For the record, these are the implementations of `bind`,`switch`, and `>=>`:

```
let bind
    (     f : 'a -> Result<'b,'c>)
    (result :       Result<'a,'c>)
    =
    match result with 
    |    Ok o -> f o
    | Error e -> Error e

let switch
   (f : 'a -> 'b)
   (x : 'a      )
   =
   f x |> Ok

let (>=>)
    (f : 'a -> Result<'b,'error>)
    (g : 'b -> Result<'c,'error>)
    =
    f >> (bind g)
```

My next attempt was using an alternative implementation for `>=>`:

```
let (>=>>)
    (f : 'a -> Result<'b,'error>)
    (g : 'b -> Result<'c,'error>)
    x
    =
    match (f x) with
    |    Ok o -> g o
    | Error e -> Error e
```

Here's the test invocation on `dotnet fsi`:

```
(id >=>> (switch ((+) 2) : int -> Result<int,string>))
((Ok 27) : Result<int,string>)
//=> Ok 29
```

I'm already stuck at why `>=>>` does not blow up on `id` as its first argument?

This is how I would think evaluation goes, but apparently this is not it: 

```
id >=>> switch ((+) 2)
          |
          V
(>=>>) id (switch ((+) 2))
          |
          V
    match (id x) with
    |    Ok o -> (switch ((+) 2)) o
    | Error e -> Error e
```

functional-programming monads monadic-composition

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